How big could it be...

First if $f(n)=\sum_{d|n} g(d)$ then $g(n) =\sum_{d|n} \mu(n/d) f(d)$ . (Mobius inversion).

We have $gcd(k,n)=\sum_{d|n,d|k} \phi(d) =\sum_{d|n} e_k(d)\phi(d)$ , such that $e_k(d)=1$ if $d|k$ and $0$ otherwise and $\phi(n)$ is Euler's totient function.

We get $g(n) =\sum_{d|n} \mu(n/d) f(d)=\sum_{k=0}^{\infty} \frac{1}{2^{k}}\sum_{d|n} gcd(k,d)\mu(n/d)=\sum_{k=0}^{\infty} \frac{1}{2^{k}}e_k(n)\phi(n)$ .( $(e_k\phi *1)(n)=gcd(k,n)$ )

Finally $g(n)=\phi(n)\sum_{k=0}^{\infty} \frac{1}{2^{kn}}=\phi(n)\frac{2^{n}}{2^{n}-1}$ . The answer will be $v_2(\phi(12321))+12321 =12324$

By Mobius inversion we have $g(n) = \sum\limits_{d|n} \mu(n/d) f(d)$ where $\mu$ is the Mobius function. In this case we have $12321 = 3^2 \cdot 37^2$ so we get $g(12321) = f(12321) - f(4107) - f(333) + f(111).$ So $g(12321) = \sum_{i=0}^\infty \frac{a_i}{2^i}$ where $a_i = \gcd(i,12321)-\gcd(i,4107)-\gcd(i,333)+\gcd(i,111).$ It's not too hard to see that $a_i$ is nonzero if and only if $12321 | i,$ in which case it equals $12321-4107-333+111=7992.$

So $g(12321) = \sum\limits_{j=0}^\infty \frac{7992}{2^{12321j}} = \frac{7992}{1-\frac1{2^{12321}}} = \frac{7992 \cdot 2^{12321}}{2^{12321}-1}.$ Since $7992 = 8 \cdot 999,$ we get $v_2(p) = v_2(7992)+12321 = \fbox{12324}.$